Question

The van der Waal's constant for a gas are a = 3.6 atm L² /mol , b = 0.6 L/mol. and take R = 0.08 expressed in same units as that of a and b. Find the Boyle's temperature of the gas. ​

Answer 1

Given

Van der Waal's constant of a gas

• a = 3.6 atm L² $mol^{-1}$

• b = 0.6 L $mol^{-1}$

To Find

Boyle's Temperature.

Concepts

$\pink{\rightarrow}$ Boyle's temperature is that temperature at low pressure region, where gas follows ideal law.

The formula to calculate Boyle's Temperature is $\boxed{\pink{ \frac{a}{Rb}}}$ .

where,

• a and b are Van der Waal's constant .

• R is the gas constant.

Calculations

We know,

Boyle's Temperature = $\frac{a}{Rb}$

putting values , we get

$T\:= \frac{3.6}{0.08\:×\:0.6}\\\\ T=\:\frac{600}{8}$

=> T = 75 K.

$\boxed{\boxed{\orange{Hence,\:the\:Boyle's\:temperature\:of\:the\:gas\:is\:75\:K}}}$